Cho dãy chuyển hóa sau: Y ← $\stackrel{+Ba(OH{)}_2}{\to }$ Z ← $\stackrel{+X}{\to }$ NaOH $\stackrel{+X}{\to }$ Y $\stackrel{+Ba(OH{)}_2}{\to }$ T.
Các chất X, Y tương ứng là

2NaOH + Ba(HCO${}_3$)${}_2$ (X) $\to$ Na${}_2$CO${}_3$ (Y) + BaCO${}_3$ + 2H${}_2$O
Na${}_2$CO${}_3$ + Ba(OH)${}_2$ $\to$ BaCO${}_3$ (T) + 2NaOH
NaOH + Ba(HCO${}_3$)${}_2$ (X) $\to$ NaHCO${}_3$ (Z) + BaCO${}_3$ + H${}_2$O
2NaHCO${}_3$ + Ba(OH)${}_2$ $\to$ BaCO${}_3$ + Na${}_2$CO${}_3$ + 2H${}_2$O